# Math 2280 - Lecture 29 Dylan Zwick Spring 2013

```Math 2280 - Lecture 29
Dylan Zwick
Spring 2013
A few lectures ago we learned that the Laplace transform in linear,
which can enormously simplify the calculation of Laplace transforms for
sums and scalar multiples of functions. The next natural question is what
relations, if any, are there for Laplace transforms of products? It might be
“nice” if
L(f (t) &middot; g(t)) = L(f (t)) &middot; L(g(t))1
But we can see right away that this would be ridiculous:
1
1
= L(1) = L(1 &middot; 1) = L(1) &middot; L(1) = 2 .
s
s
That ain’t right. So, we must state that, in general
L(f (t) &middot; g(t)) 6= L(f (t)) &middot; L(g(t)).
So, that doesn’t work, and the question remains. Namely, what, if any,
relation is there between products and Laplace transforms? It turns out
there is an operation called the convolution of two functions, and this operation will give us the best answer to our question.
Today’s lecture corresponds with section 7.4 of the textbook, and the
assigned problems are:
Section 7.4 - 1, 5, 10, 19, 31
1
This is NOT true. Not true, not true, not true. Don’t memorize this because it’s NOT
TRUE.
1
Derivatives, Integrals, and Products of Transforms
need to define an operator called convolution. The convolution of two functions f (t) and g(t) is denoted by f (t) ∗ g(t), and is defined by:
f (t) ∗ g(t) =
Z
t
f (τ )g(t − τ )dτ .
0
We note that it’s a straightforward if tedious task to verify that the operator is commutative and associative.
Example - Calculate the convolution:
t2 ∗ cos t.
2
More room for the example problem.
Now, why are these convolutions important? Well, as you might expect, they relate to products of Laplace transforms in a big way. This relation is given by the next theorem.
Theorem - Suppose that f (t) and g(t) are piecewise continuous for
t ≥ 0 and that |f (t)| and |g(t)| are bounded by Mect as t → ∞. Then
the Laplace transform of f ∗ g exists for s &gt; c, and
L(f (t) ∗ g(t)) = L(f (t)) &middot; L(g(t)).
3
Example - Calculate the inverse Laplace transform of:
F (s) =
1
.
+ 1)
s(s2
Finally, just as there are relations for the Laplace transform of a derivative, there are relations for the derivative of a Laplace transform. These
are:
L(t &middot; f (t)) = −F ′ (s),
and in greater generality
L(tn &middot; f (t)) = (−1)n F (n) (s).
Also, conversely
Z ∞
f (t)
=
F (σ)dσ.
L
t
s
Deriving these relations is fairly straightforward, and is done at the
end of section 7.4 from the textbook.
4
Example - Calculate the Laplace transform:
L(t sin (3t)).
Example - Calculate the Laplace transform:
L
e3t − 1
.
t
5
```